Is the maximum bound of Euclidean distance between two probability distributions equal to $\sqrt{2}$? I used Euclidean distance to compute the distance between two probability distribution. The example of computation shown in the Figure below.

As my understanding, the maximum distance occur while $A_1 = 0$ and $B_1 = 1$, $A_2 = 1$ and $B_1 = 0$. Whereas the Euclidean distance will be $$d = \sqrt{ 1^2 + 1^2 }$$
Is anyone have the theorem and mathematically prove about this? and does my assumption is correct, that $\sqrt2$ is the maximum bound for this distance? 
Another question, while the bins more than two/ many bins, is that $\sqrt 2 $ still the maximum bound? 
 A: $d_{xy}^2 = \sum{(x-y)^2} = \sum x^2 + \sum y^2 - 2\sum xy$.
Given that in probability vectors all values are nonnegative, $d^2$ is max when the last term is zero. Then $d^2 = \sum x^2 + \sum y^2$.
In a probability vector all values (which are between 0 and 1) sum up to 1, $\sum x = \sum y = 1$. In such a vector, its theoretical maximum of $\sum v^2$ is attained when all its entries are 0 except one which is 1, and this maximum is $1 = \sum v$. Thus the theoretically maximal squared distance between two such vectors is 2: it is when $\sum x^2 = \sum x$ and $\sum y^2 = \sum y$. It also follows from the above description, that then $\sum xy$ can very easily happen to be zero (since in each vector there is just single nonzero element).
A: Yes, in the 2 category case, $\sqrt{2}$ is the maxima achieved at both $A=(1,0), B=(0,1)$ and $A=(0,1), B=(1,0)$.
I'll supply a heuristic argument which should be straightforward to rigorously demonstrate. Define $d_1 = A_1 - B_1$ and $d_2 = A_2 - B_2$. Then euclidean distance can be thought of as circular contours around the origin.

Furthermore, we know the properties of probability distributions dictate that $A_1 + A_2 = 1$ and $A_i \in [0,1]$ (same for $B$). This implies that $d_1 = -d_2$, meaning our solution space is the line from the upper left of the contour plot at $(-1,1)$ to the lower right at $(1,-1)$. Given that euclidean distance contours are concentric circles w.r.t. $d_1$ and $d_2$, the maxima are the points furthest away from the origin, $d_1 = 1, d_2 = -1$ and $d_1 = -1, d_2 = 1$.
It follows from the previous constraints on $A$ and $B$ that these points uniquely occur at $A=(1,0), B=(0,1)$ and $A=(0,1), B=(1,0)$ respectively.
Edit, since OP is looking for a proof. I'm not sure of a well known result, but a proof of the 2 category case follows.
Using $d_1=-d_2$, we can redefine euclidean distance as $e = \sqrt{d_1^2 + -d_1^2}$. Differentiating, we find $\frac{\partial e}{\partial d_1} = \frac{2d}{\sqrt{d^2}}$. This implies that $e$ is maximized at maximal $|d_1|$.
To find maximal $|d_1|$, consider that it's constrained by the line $A_1=1-A_2$ in the region $A_1 \in [0,1]$. Our partial derivatives of $d_1$ are $\frac{\partial d_1}{\partial A_1} = 1$ and $\frac{\partial d_1}{\partial B_1} = -1$, so there's no stationary points, implying the minima & maxima happen on the boundaries. The first boundary, along the constrained line, is $A_1 = 0$ and $B_1 = 1$ with value $d_1 =-1$, and the other boundary point is $A_1 = 1$ and $B_1=0$ with a value of $d_1 = 1$. These are are maxima and minima of $d_1$, both of which maximize $|d_1|$, uniquely determine $d_2$, and as shown above maximize $e$.
